Volumetric Analysis - Online Tutor, Practice Problems & Exam Prep

Concentration of solutions can be expressed through weight percent, volume percent, and weight volume percent. Weight percent is calculated as the mass of solute divided by the mass of the solution, multiplied by 100. Volume percent uses the volume of solute over the volume of solution, also multiplied by 100. Weight volume percent combines both, using the mass of solute in grams over the volume of solution in milliliters, multiplied by 100. Understanding these concepts allows for interconversion with molarity and molality, enhancing comprehension of solution chemistry.

Volumetric Analysis Concept 1

Now in addition to molarity and molality, we can express the concentration of solutions in a few other ways. So here we have 3 formulas: weight percent, volume percent, and weight volume percent. With weight percent, which is also called mass percent, we have weight over weight or mass over mass. That'll equal the mass of our solute divided by the mass of our solution times 100. On the other side of this, we have volume percent.

Whereas weight or mass percent looks at mass, volume percent is going to look at volume. So we'd have volume over volume here. In this case, it'd be the volume of the solute, usually in mL, divided by the volume of our solution, also in mL, and it'd be times 100. And then here we have weight volume percent, which is kind of a mixing of these 2. So here it's a weight over volume type of situation.

And we're going to say here when it comes to weight, we're looking at the mass of our solute in grams divided by the volumes of our solution, usually milliliters times 100. Now here if we go back to weight percent, this typically is grams over grams as well. So just remember, we have these three other ways of looking at the concentration of a given solution, and they themselves can connect to molarity and molality. Through different types of manipulation, we can go between these 5 different types of ideas. So pay attention as we cover different types of problems that really ask us to interconvert between these 5 different ideas.

Volumetric Analysis Example 1

Here in this example question, it states that a 5.12 liter sample of a solution contains 0.300 grams of potassium sulfate. Determine the weight percent of potassium sulfate if the density of the solution is 1.30 grams per milliliter. We're also given the molecular weight of potassium sulfate as 174.26 grams per mole. Alright. So we need to find the weight percent or mass percent.

We'd say here that the weight percent will equal the mass of our solute, in this case, potassium sulfate. We already have that portion, 0.300 grams of potassium sulfate, and it's going to be divided by the mass of our solution. So, grams of solution, and then we'd multiply that by 100. Now here, if we look at the information given to us, we have a 5.12-liter sample of the solution.

We have the volume of the solution, and we're given the density of our solution. We can use these two together to help isolate the grams of our solution. So what we do here is we have a 5.12-liter solution. We're going to say here that 1 liter is equal to 1000 milliliters. You could also say 1 milliliter is equal to 10^-3 liters, either way is fine.

Since we have milliliters of solution now, we can bring in the density, which is 1.30 grams of solution per 1 milliliter of solution. The milliliters cancel out, and now we have grams of solution at the end, which comes out to 6656 grams of solution. We can take that and plug it in. When we plug this into our calculators, we get at the end = 0.300 g 6656 g × 100 Percent=3.456 × 10^-3 percent. If we look at the number of significant figures within the question, we have three significant figures throughout.

So at the end, we'd write 3.46 × 10^-3 percent. This would be our final answer.

Parts per million of Lead

Alright. So here it says, when lead levels in blood exceed 0.80 parts per million, the level is considered dangerous. We're told that 0.80 parts per million means that 1,000,000 grams of blood contain 0.80 grams of lead. We're also told that given the density of blood is 1,060.0 kilograms per meter cubed, how many grams of lead would be found in 550 milliliters of blood with a lead level of 0.583 parts per million. This one is a big jumbled mess of numbers and expressions.

Let's organize things out a little bit. First of all, they want us to figure out the grams of lead. So that's what we're looking for, grams of lead. Now, they're telling us a lot of different things. We're accustomed to saying that parts per million means milligrams per 1 liter, but here we don't need the molarity of the solution.

Plus, they tell me here that in this case, we're going to say that 0.80 parts per million means 1,000,000 grams of blood corresponds to 0.80 grams of lead. So they're giving us a conversion factor to use. Now they're saying that the density is this value here. So this is the density of my blood solution for every 1 meter cubed of blood. We have 550 milliliters of blood.

And they gave me this as a reference point on how they want me to manipulate parts per million. The actual parts per million that we're dealing with is this one here, 0.583 parts per million. Based on what they've told me for the 0.80 parts per million, this really means 0.583 grams of lead for every 1,000,000 grams of blood. All we have to do is realize that the grams of lead that I need to isolate are right here which means I have to get rid of all these other units. I have to cancel out kilograms of blood, meters cubed, milliliters of blood and the other grams of blood on the bottom, and be left with just the grams of lead.

We're going to start out first with just the 550 mL of blood because it's easy to manipulate 1 unit instead of things mixed in with multiple units. So we bring down the 550 mL of blood. This is volume. What I'm going to do first is I'm going to convert the 1 meter cubed that we have on the bottom here to centimeters cubed. So that way they can cancel out with these milliliters here.

So we have 1 meter cubed. We're going to say here, meters go on the bottom, centimeters go on top. 1 centi is 10 to the negative 2. I'm going to cube this entire thing so that the meters cubed can cancel out. But what does cubing really mean?

Well, if I was to rewrite this, it would really mean that I have 1 meter cubed times 1 centimeter cubed. Now, when I cube this power, it just really means that the negative two is multiplying with the 3. So it'd be 10 to the negative six here at the bottom. So it's 10 to the 6 centimeters cubed that will be here on the bottom now. Again, the reason I'm changing to centimeters cubed is that milliliters and centimeters cubed are the same thing.

So now these 2 units can cancel one another out. Okay, so centimeters cubed, milliliters both cancel each other out. Now I have kilograms of blood. Next, I need to change kilograms of blood into grams of blood, and I'm doing that so that I can cancel out with these grams of blood later on.

So 1 kilogram is equal to 1,000 grams. So kilograms cancel out. Now I have grams of blood. We're going to have 1,000,000 grams of blood on the bottom, and we have 0.583 grams of lead on top. So grams of blood cancels out, and what I'm left with is grams of lead, which is what I was looking for.

When you punch that into your calculator, you get \(3.39889 \times 10^{-4}\) grams of lead based on significant figures. So the calculations that we're using for our calculations, this one here has 5 significant figures, this one here has 5 significant figures and this one here has 3 significant figures. So we want our answer at the end to have 3 significant figures so this comes up to \(3.40 \times 10^{-4}\) grams of lead as my final answer. This word problem had a lot of values being thrown at us, but again, the approach should always be to write down what you're looking for first and then once you've done that, write down all the given information on the other side.

What you have to find as your answer can be located in some way within the given information. You just have to properly organize things and then see how things connect and cancel out to give you the final units that you want at the end.

Percent Composition

In the following example, we're told an 8.13% aluminum sulfate solution has a measured density of 1.235 grams per milliliter. Calculate the molar concentration of sulfate ions in the solution. Now realize here that we don't want the molarity of the aluminum sulfate solution. We just want the molarity of the sulfate ions themselves. So here we're looking for the molarity of sulfate ions.

So that would just be the moles of sulfate ions divided by the liters of solution. What information are we given? Well, we're told that we have 8.13% of aluminum sulfate solution. So that 8.13%, what does that really mean? Well, that means that we have 8.13 grams of aluminum sulfate for every 100 grams of solution.

We're also told that the density of the solution is 1.235 grams of solution for every 1 milliliter of solution. In addition, we're given the mass of aluminum sulfate. But with these types of calculations, we need that. We can write that down too. We'll incorporate it within our calculations.

These are the two pieces of information that we're being given and with this, we have to figure out the molarity of sulfate ions. What we're going to do here is we're going to start out by bringing down the mass percent of aluminum sulfate solution. We have 8.13 grams of aluminum sulfate, and that's over 100 grams of solution. What we're going to do first is we're going to convert the grams of aluminum sulfate into moles of aluminum sulfate by using the molecular mass or molecular weight they were given within the question. We want to get rid of grams so it goes on the bottom.

So we have 342.17g of aluminum sulfate on the bottom. And then we have 1 mole of aluminum sulfate on top. What happens here is this cancels out with this. The reason we had to figure out the moles of aluminum sulfate is that once we have the moles of the entire compound, we can do a mole to mole comparison to find the moles of just sulfate ions. At this point, we're going to say that for every one mole of the entire aluminum sulfate compound, there are exactly 3 moles of sulfate ions.

So that's 3 moles of sulfate ions. So the moles cancel out. At this point we have moles of sulfate ion over grams of solution. We're not done yet because we need liters of solution on the bottom. Now, how do I get rid of those grams of solution?

Well, we have the density of our solution there. We're going to bring that down. We want to cancel out these grams of solution here on the bottom. So we're going to have the grams of solution of the density on top.

And then, 1 milliliter of solution on the bottom. So grams of solution cancel out. Now I have milliliters of solution on the bottom. I want to cancel out those ml, so I'm going to put 1,000 ml of solution on top and t

Percent Composition

Alright. So here it says, this density of a 33.8% solution of sodium acetate is 1.10 grams per milliliter. It says a reaction requires 68.8 grams of sodium acetate. What volume of the solution do you need if you want to use a 50% excess of sodium acetate? Alright.

So this question is a bit different from what we have seen in the past. We're dealing here with an excess of a particular compound. But before we talk about that excess, let's see what they are asking us to find. They want us to find the volume of the solution. So here we're looking for the milliliters.

I'm just going to say milliliters as my default amount or units, milliliters of solution. Next, we're going to write down all the given information, and from that, we'll solve for the milliliters of solution. All right. So, we're told that we have a 33.8% solution, so that means we have 33.8 grams of sodium acetate, divided by 100 grams of solution. We're told that we have a density of solution that's 1.10 grams per milliliter.

So that's 1.10 grams of solution per 1 milliliter of solution. And we can see here that the milliliters of solution I want to isolate, we can find them right here. So we know that we're going to end with a density of solution in some way to get milliliters of solution at the end. Also, they're telling us that we require 68.8 grams of sodium acetate, but that we want a 50% excess of that. So what exactly does that mean?

Well, in a reaction, when we're talking about theoretical yield, theoretical yield represents 100 percent. What we want now is we want an additional 50%. So we want 150%. But when it comes to percentages, we don't use them in our calculations that way. What we're going to do is we're going to divide this by 100, so that's going to come out to 1.50.

We're going to take that 1.50 and multiply it by the 68.8 grams of sodium acetate. That'll give us our 50% excess that we need. So multiply it by 1.50, so that gives me 103.2 grams of sodium acetate. Now, we know that we want to isolate milliliters of solution. We're going to start out by using first grams of sodium acetate because it's the easiest to manipulate because it's just one unit.

So we have 103.2 grams of sodium acetate. These grams of sodium acetate can cancel out with these grams of sodium acetate. So we're going to put on the bottom 33.8 grams of sodium acetate and 100 grams of solution on top. Now, we want to cancel out those grams of solution. So take the density, we take the 1.10 grams of solution, put them on the bottom, and then the 1 milliliter of solution on top.

So grams of solution cancel out, so at the end I'll have milliliters of solution. This comes out to 277.569 milliliters of solution. Here, this has 3 significant figures, 3 significant figures, 3 significant figures. 50%, that has one significant figure but we'll go with the other ones. That gives me 278 milliliters of my solution.

Okay. So that's my answer there. So we're accustomed to seeing these types of word problems. The new twist was looking at excess. Realize that when we're talking about theoretical yield, that's 100%.

So any excess would just be that number added to the 100%. We would divide that new percentage by 100 to get its decimal form and then multiply it by the theoretical yield which in this case was 68.8 grams of sodium acetate. Doing that helped us to isolate the volume of my solution in the end. Now we've seen this one. Try to attempt the practice question left here on the bottom.

Again, don't worry if you get stuck. Just come back and see how I approach that practice problem left at the bottom of the page.

Determine the number of mmoles of sulfuric acid that would be found within a 120 g sample that is 79.9% sulfuric acid.

• A. 0.960 mmol 
• B. 980 mmol
• C. 0.33 mmol
• D. 1.7 mmol

Dilution

So we're going to say here a dilution involves the addition of water to a concentrated solution. We're going to say typically a given volume of a concentrated solution is placed in a volumetric flask and water is added to a determined mark. Now typically when we're talking about dilutions, we're going to use the dilution equation which is m1V1=m2V2. Here, m1 represents the concentrated molarity. This is basically our concentration of molarity before water has been added.

V1 represents our initial volume, so before any additional water has been added. m2 represents our diluted concentration and V2 represents our final volume after we've added the water. Now we're going to say typically that m1, which is more concentrated, will be the larger concentration than m2. And V2 is the larger volume than V1. V2 represents your final volume which equals your initial volume which is V1 plus volume of added water.

So just remember, anytime we're adding water to any type of solution, that represents a dilution which means that we may use this formula in some way or another. Knowing this will help guide us to the exact way of solving the following examples that are given on the bottom. You can take a look at example 1, and attempt it on your own. But if you get stuck, just go on to the next video and see how I approach answering the example question given below.

Dilutions

Here it's asking how many grams of 53.1 weight percent sodium chloride should be diluted to 2.50 liters to make 0.15 molar NaCl. Now in this question, we're not dealing with m1v1 = m2v2 because we only have one volume and one molarity involved within the question. What this question itself is asking, it's asking us for the grams of solution. So how many grams of solution are needed based on what they've told us. Now with this information, we're going to put all the given information on the other side and organize it to help isolate our grams of solution. So we have 53.1 weight percent sodium chloride solution. So that means we have 53.1 grams of sodium chloride over one hundred grams of solution. There goes the grams of solution we know we need to isolate at the very end of all our calculations. We're told that we have 2.50 liters. Really mean? Well, molarity is moles over liters, so 0.15 molar really means that I have 0.15 moles of sodium chloride per 1 liter of solution. So we kno

If 920 mL of water is added to 78.0 mL of a 1.28 M HBrO₄ solution what is the resulting molarity? 

• A. 0.110 M
• B. 15.1 M
• C. 0.100 M
• D. 0.119 M

Volumetric Analysis Example 6

Here we're told that a student prepared a stock solution by dissolving 25 grams of sodium hydroxide in enough water to make a 150 ml of solution. The student took 20 ml of the stock solution and diluted it with enough water to make 250 ml of solution. And then finally, taking 75 ml of that solution and dissolving it in water to make 500 ml of solution. Now what is the concentration of sodium hydroxide for this final solution? We're also given the molecular weight of sodium hydroxide as being 40 grams per mole.

Now this question here represents a serial dilution. Basically, we make our original concentration, which is our stock solution, and we're going to go through a series of dilutions to get to our final molarity at the end. So our first step is to determine what our stock solution concentration is. We have 25 grams of sodium hydroxide. For every one mole of sodium hydroxide, it is 40 grams.

So this adds up to 0.625 moles of sodium hydroxide. We're doing this beca

Dilutions

The density of 63.7 weight percent sodium hydroxide is 0.915 grams per milliliter. How many milliliters of water should be diluted to 850 ml's to create 0.425 molar sodium hydroxide. Here, they're asking us how many ml's of water. It's really asking how many ml of solution should be diluted. That's what we're looking for.

What we're going to do now is we're going to isolate all the given information and see how we can isolate those ml's of solution. We're told here that we have 63.7 grams of sodium hydroxide for every 100 grams of solution, reflecting a 63.7 weight percent sodium hydroxide. We're told that the density of the solution is 0.915, so that's 0.915 grams of solution per 1 ml of solution. What else are we given? We're told that we have 850 ml to create 0.425 molar sodium hydroxide.

So, we have 850 ml. And then remember, molarity is moles over liters, so that's just 0.425molNaOH&

Here’s what students ask on this topic:

What is the difference between molarity and molality?

Molarity (M) is defined as the number of moles of solute per liter of solution. It is expressed as:

$M=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{liters} \mathrm{of} \mathrm{solution}}$

Molality (m) is defined as the number of moles of solute per kilogram of solvent. It is expressed as:

$m=\frac{\mathrm{moles} \mathrm{of} \mathrm{solute}}{\mathrm{kilograms} \mathrm{of} \mathrm{solvent}}$

The key difference is that molarity depends on the volume of the solution, which can change with temperature, while molality depends on the mass of the solvent, which remains constant regardless of temperature changes.

How do you calculate weight percent in a solution?

Weight percent (also known as mass percent) is calculated by dividing the mass of the solute by the mass of the solution and then multiplying by 100. The formula is:

$\mathrm{Weight} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 5 grams of solute in 100 grams of solution, the weight percent would be:

$\frac{5}{100} \times 100=5\%$

What is volume percent and how is it calculated?

Volume percent is a way to express the concentration of a solution, calculated by dividing the volume of the solute by the volume of the solution and then multiplying by 100. The formula is:

$\mathrm{Volume} \mathrm{percent}=\frac{\mathrm{volume} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 10 mL of solute in 100 mL of solution, the volume percent would be:

$\frac{10}{100} \times 100=10\%$

How do you convert between molarity and weight percent?

To convert between molarity (M) and weight percent, you need the density of the solution and the molar mass of the solute. The steps are:

1. Calculate the mass of the solute using molarity and volume of the solution.

2. Determine the total mass of the solution using its density.

3. Use the weight percent formula:

$\mathrm{Weight} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{mass} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have a 1 M solution of NaCl (molar mass = 58.44 g/mol) and the density of the solution is 1.05 g/mL, for 1 L of solution:

Mass of NaCl = 1 mol × 58.44 g/mol = 58.44 g

Mass of solution = 1 L × 1.05 g/mL × 1000 mL/L = 1050 g

Weight percent = (58.44 g / 1050 g) × 100 ≈ 5.56%

What is weight volume percent and how is it used?

Weight volume percent (w/v%) is a concentration measure that combines the mass of the solute and the volume of the solution. It is calculated by dividing the mass of the solute (in grams) by the volume of the solution (in milliliters) and then multiplying by 100. The formula is:

$\mathrm{Weight} \mathrm{volume} \mathrm{percent}=\frac{\mathrm{mass} \mathrm{of} \mathrm{solute}}{\mathrm{volume} \mathrm{of} \mathrm{solution}} \times 100$

For example, if you have 5 grams of solute in 100 mL of solution, the weight volume percent would be:

$\frac{5}{100} \times 100=5\%$

This measure is commonly used in medical and biological contexts, such as preparing solutions for intravenous fluids.